Assume that boys and girls are equally likely

Approach 1

A random variable ?X ?is said to be binomially distributed if its probability mass function (PMF) is given by

p(x)=x!(n−x)!n!?px(1−p)n−x ?for x=0,1,2,...,n?

where ?n ?is the number of trials (or sample size) and ?p ?is the probability of success.

In detail, binomial experiment possesses the following properties: 

• the experiment consists of n identical trials
• each trial results in one of two outcomes: success or failure
• the probability of success on a single trial is equal to p and remains the same from trial to trial while the probability of failure is equal to q=1-p
• the trials are independent

For the problem, we let X = number of girls in three children. We know that

n = 3

p = 0.5 (since the probability that a child is a girl is 0.5 (equally likely to the probability that a child is a boy))

So the PMF would be

p(x)=x!(3−x)!3!?0.5x(1−0.5)3−xfor x=0,1,2,3

Now we write the probability of exactly 1 girl out of three children as P(X=1). Thus

P(X=1)=p(1)

P(X=1)=1!(3−1)!3!?0.51(1−0.5)3−1

P(X=1)=0.375

Approach 2
We know that the following combinations are

bbb, bbg, bgb, gbb, bgg, gbg, ggb, bbb

The sample points where there is exactly 1 girl are

bbg, bgb, gbb

So the probability would be total number of outcomesnumber of exactly 1 girl?=83?=0.375